For all positive real numbers x, (xˆ2)-x≥0.

I don't think this is true since for instance x can equal 2/3 and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong?

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- Sep 14th 2011, 06:36 PMAquameatwadQuick Proof on real numbers
For all positive real numbers x, (xˆ2)-x≥0.

I don't think this is true since for instance x can equal 2/3 and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong? - Sep 14th 2011, 06:43 PMSudharakaRe: Quick Proof on real numbers
- Sep 14th 2011, 06:50 PMAquameatwadRe: Quick Proof on real numbers
I see, so is the original proof "For all positive real numbers x, (xˆ2)-x≥0. " is only correct if you set the parameter x≥1.

So the original statement is false since 2/3 is a positive real number, and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Correct? Just want to make sure whats in my head is correct. - Sep 14th 2011, 08:26 PMSudharakaRe: Quick Proof on real numbers
- Sep 14th 2011, 08:51 PMProve ItRe: Quick Proof on real numbers
If you want to prove this directly, you could complete the square...

$\displaystyle \displaystyle \begin{align*} x^2 - x &\geq 0 \\ x^2 - x + \left(-\frac{1}{2}\right)^2 &\geq \left(-\frac{1}{2}\right)^2 \\ \left(x - \frac{1}{2}\right)^2 &> \frac{1}{4} \\ \left|x - \frac{1}{2}\right| &> \frac{1}{2} \\ x - \frac{1}{2} \leq -\frac{1}{2} \textrm{ or }x - \frac{1}{2} &\geq \frac{1}{2} \\ x \leq 0 \textrm{ or } x &\geq 1 \end{align*}$

So the inequality $\displaystyle \displaystyle x^2 - x \geq 0$ is only true when $\displaystyle \displaystyle x \leq 0$ or $\displaystyle \displaystyle x \geq 1$. - Sep 15th 2011, 04:05 AMHallsofIvyRe: Quick Proof on real numbers
S0mewhat simpler, I think:

$\displaystyle x^2- x= x(x- 1)\ge 0$ if and only if both factors on the left have the same sigh. If $\displaystyle x\ge 1$, x> 0 and $\displaystyle x-1\ge 0$, so that is true. If $\displaystyle x\le 0$, x< 1 so that is true. But if $\displaystyle 0< x< 1$, x is positive while x- 1 is negative so it is not true.